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2 years ago

Which element has chemical properties that are most similar to the chemical properties of sodium?

Chemistry

1 answer:

notka56 [123]2 years ago

7 0

Answer:

Lithium

Explanation:

So, the elements with similar properties to sodium (Na) are all of those elements in the same Group. The ones "most" similar would be the ones closest in mass as well. Those would be Lithium (Li) and Potassium (K)

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Which of the following groups of chemical elements would you expect to be most alike in their physical and chemical properties?

juin [17]

Many elements show very strong similarities to each other.<span>For example, lithium (Li), sodium (Na), and potassium (K) are all soft, very reactive metals.

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8 0

2 years ago

Complete the equation for the ionization of water by drawing the conjugate acid and conjugate base. Include lone pairs of electr

kodGreya [7K]

Answer:

Here's what I get

Explanation:

I followed the instructions and got the diagram below.

5 0

3 years ago

Read 2 more answers

5. You are investigating an arson scene and you find a corpse in the rubble, but you suspect that the victim did not die as a re

Marina CMI [18]

Answer:

See the answer below.

Explanation:

Fire has three major components:

Heat
Smoke
Gases ( in form of CO, CO2 etc)

If the victim had died as a result of the fire, he/he would have inhaled smoke and hot gases from the fire. These components would have resulted in traces of burns and soot deposition in the trachea and lungs as well as traces of CO in the blood of the victim.

If the analysis of the victim's corpse does not reflect some of the results above, it can be effectively concluded that the victim has been dead before the fire.

<em>The single most important indicator of death by the fire would be the presence of CO in the blood of the victim's corpse. All others might be to a less significant degrees.</em>

8 0

2 years ago

Calculate the amount of heat released when 27.0 g H2O is cooled from a liquid at 314 K to a solid at 263 K. The melting point of

BlackZzzverrR [31]

Answer : The amount of heat released, 45.89 KJ

Solution :

Process involved in the calculation of heat released :

$(1):H_2O(l)(314K)\rightarrow H_2O(l)(273K)\\\\(2):H_2O(l)(273K)\rightarrow H_2O(s)(273K)\\\\(3):H_2O(s)(273K)\rightarrow H_2O(s)(263K)$

Now we have to calculate the amount of heat released.

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+\Delta H_{fusion}+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

Q = amount of heat released = ?

m = mass of water = 27 g

$c_{p,l}$ = specific heat of liquid water = 4.184 J/gk

$c_{p,s}$ = specific heat of solid water = 2.093 J/gk

$\Delta H_{fusion}$ = enthalpy change for fusion = 40.7 KJ/mole = 40700 J/mole

conversion : $0^oC=273k$

Now put all the given values in the above expression, we get

$Q=[27g\times 4.184J/gK\times (314-273)k]+40700J+[27g\times 2.093J/gK\times (273-263)k]$

$Q=45896.798J=45.89KJ$ (1 KJ = 1000 J)

Therefore, the amount of heat released, 45.89 KJ

7 0

3 years ago

Read 2 more answers

If a solution containing 45.101 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of s

AnnyKZ [126]

Answer:

14.533 grams of solid precipitate of mercury(II) dichromate will form.

Explanation:

$Hg(CH_3COO)_2(aq)+Na_2Cr_2O_7(aq)\rightarrow HgCr_2O_7(s)+2CH_3COONa(aq)$

Moles of mercury(II) acetate = $\frac{45.101 g}{318.70 g/mol}=0.14152 mol$

Moles of sodium dichromate = $\frac{12.026 g}{261.97 g/mol}=0.045906 mol$

According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) acetate

This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.

According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) dichromate

Mass of 0.045906 moles of mercury(II) dichromate:

0.045906 mol × 316.59 g/mol = 14.533 g

14.533 grams of solid precipitate of mercury(II) dichromate will form.

3 0

3 years ago